Can we define tensor product of modules over an algebra? 
Let $A$ be an algebra over a commutative ring $k$ and $M$ and $N$ be modules over $A.$ Is there any natural way to define tensor product of $M$ and $N$ over the algebra $A\ $?

My idea is that since $A$ is an algebra it already has the ring structure. So we can consider the tensor product of $M$ and $N$ when $A$ is regarded as a ring. But when $A$ is a Hopf algebra the monoidal structure is defined in terms of the comultiplication. Namely, given $A$-modules $M$ and $N,$ the $A$-module structure on $M \otimes N$ is defined as follows $:$
$$a \cdot (m \otimes n) : = \Delta (a) \cdot (m \otimes n) = \sum\limits_{a} a_{(1)} m_1 \otimes a_{(2)} m_2$$
by using Sweedler's notation. But why do we need to use the comultiplication here? Can't we simply use the bilinearity of the tensor products? I am pretty confused at this stage. Any suggestion in this regard would be greatly appreciated.
Thanks for your time.
 A: I think we need to distinguish between at least three (or four) different kinds of “tensor product”.
1 ♦ Tensor products over commutative rings
Let $$ be a commutative ring (often a field) and let $M$ and $N$ be two left $$-modules.
We can then form the tensor product $M ⊗_{} N$.
This tensor product is again a left $$-modules,
As a $$-module, it is generated by elements $m ⊗ n$ with $m ∈ M$ and $n ∈ N$, subject to the following relations:
\begin{gather*}
 (m_1 + m_2) ⊗ n = m_1 ⊗ n + m_2 ⊗ n \,,
 \quad
 m ⊗ (n_1 + n_2) = m ⊗ n_1 + m ⊗ n_2 \,,
 \\
 (λ m) ⊗ n = λ ⋅ (m ⊗ n) \,,
 \quad
 m ⊗ (λ n) = λ ⋅ (m ⊗ n) \,.
\end{gather*}
The action of a scalar $λ$ on a simple tensor $m ⊗ n$ is given by
$$
 λ ⋅ (m ⊗ n) = (λm) ⊗ n = m ⊗ (λn) \,.
$$
The expression $m ⊗ n$ is thus $$-bilinear in $m$ and $n$.
2 ♣ Tensor product over arbitrary rings, naive approach
Let now $R$ be an arbitrary ring, and let $M$ and $N$ be two left $R$-modules.
We could try to mimic to above construction to define a kind of “tensor product” $M \mathbin{\widetilde{⊗}}_R N$.
We would define $M \mathbin{\widetilde{⊗}}_R N$ as the left $R$-module generated by elements $m ⊗ n$ with $m ∈ M$ and $n ∈ N$ subject to the following relations:
\begin{gather*}
 (m_1 + m_2) ⊗ n = m_1 ⊗ n + m_2 ⊗ n \,,
 \quad
 m ⊗ (n_1 + n_2) = m ⊗ n_1 + m ⊗ n_2 \,,
 \\
 (r m) ⊗ n = r ⋅ (m ⊗ n) \,,
 \quad
 m ⊗ (r n) = r ⋅ (m ⊗ n) \,.
\end{gather*}
The action of a scalar $r$ on a simple tensor $m ⊗ n$ is again given by
$$
 r ⋅ (m ⊗ n) = (rm) ⊗ n = m ⊗ (rn) \,.
$$
The expression $m ⊗ n$ is thus $R$-bilinear in $m$ and $n$.
However, this kind of tensor product has some unwanted properties.
We can observe that
\begin{align*}
 rs ⋅ (m ⊗ n)
 &=
 r ⋅ (s ⋅ (m ⊗ n))
 \\
 &=
 r ⋅ (m ⊗ (sn))
 \\
 &=
 (rm) ⊗ (sn)
 \\
 &=
 s ⋅ ((rm) ⊗ n)
 \\
 &=
 s ⋅ (r ⋅ (m ⊗ n))
 \\
 &=
 sr ⋅ (m ⊗ n)
\end{align*}
for all scalars $r, s ∈ R$, and every simple tensor $m ⊗ n$.
Every element of $M \mathbin{\widetilde{⊗}}_R N$ is a sum of simple tensors, so it further follows that the elements $rs$ and $sr$ act the same way on every element of $M \mathbin{\widetilde{⊗}}_R N$.
In other words, the module $R$-module $M \mathbin{\widetilde{⊗}}_R N$ completely ignores the (possible) non-commutativity of $R$!
One can make this “forgetting non-commutativity” a bit more precise as follows:
Let $I$ be the two-sided ideal in $R$ generated by all elements of the form $rs - sr$ with $r, s ∈ R$.
This ideal is called the commutator ideal of $R$.
The quotient ring $R / I$ and commutative, and in fact the largest possible commutative quotient of $R$.
The two left $R$-modules $M$ and $N$ result in the left $R/I$-modules $M / IM$ and $N / IN$.
We can thus form the tensor product $(M / IM) ⊗_{R / I} (N / IN)$ in the sense of section 1.
It then holds that
$$
 M \mathbin{\widetilde{⊗}}_R N ≅ (M / IM) ⊗_{R / I} (N / IN)
$$
as left $R$-modules (and equivalently, as left $(R / I)$-modules).
We see from this isomorphism that forming the tensor product $M \mathbin{\widetilde{⊗}}_R N$ amounts to killing off all non-commutativity in $R$, $M$ and $N$, and then forming the tensor product over the commutative ring $R / I$.
This is typically not what one wants when working with non-commutative rings!
3 ♥ Tensor product over arbitrary rings, better approach
The problem in the previous section occurred because the simple tensors $m ⊗ n$ were constructed to be bilinear in $m$ and $n$.
Indeed, the problem is that in the relation $(r m) ⊗ n = m ⊗ (r n)$, we are swapping the order of the terms $r$ and $m$.
This works okay in a commutative setting, but not so well in a non-commutative one.
To fix this problem, we can instead consider the relation
$$
 (m r) ⊗ n = m ⊗ (r n) \,.
$$
To make sense of this relation, we need $M$ to be a right $R$-module, and $N$ a left $R$-module.
So let $M$ be a right $R$-module and $N$ a right $R$-module.
The tensor product $M ⊗_R N$ the abelian group generated by elements $m ⊗ n$ with $m ∈ M$ and $n ∈ N$ subject to the relations
\begin{gather*}
 (m_1 + m_2) ⊗ n = m_1 ⊗ n + m_2 ⊗ n \,,
 \quad
 m ⊗ (n_1 + n_2) = m ⊗ n_1 + m ⊗ n_2 \,,
 \\
 (m r) ⊗ n = m ⊗ (r n) \,.
\end{gather*}
The expression $m ⊗ n$ is not $R$-bilinear in $m$ and $n$.
Instead, it is “$R$-balanced”.
However, this kind of tensor product of $R$-modules (namely of a right-module with a left-module) is not again any kind of $R$-module!
This probably seems strange it first.
But we still have the following construction, which is extremely useful in praxis:

*

*If $M$ is not just a right $R$-module, but an $S$-$R$-bimodule for some other ring $S$, then $M ⊗_R N$ becomes a left $S$-module via
$$
 s ⋅ (m ⊗ n) = (s m) ⊗ n \,.
$$

*If $T$ is not just a left $R$-module, but an $R$-$T$-bimodule for some other ring $T$, then $M ⊗_R N$ becomes a right $T$-module via
$$
 (m ⊗ n) ⋅ t = m ⊗ (n t) \,.
$$

*We can also combine both of the above:
if $M$ is an $S$-$R$-bimodule and $N$ is an $R$-$T$-bimodule, then $M ⊗_R N$ becomes an $S$-$T$-bimodule via
$$
 s ⋅ (m ⊗ n) ⋅ t = (sm) ⊗ (nt) \,.
$$
So the tensor product $M ⊗_R N$ can still have useful module structures, if only the original modules $M$ and $N$ have suitable bimodule structures.
Comparing to section 1
The first notion of tensor products, over a commutative ring, is actually a special case of this general notion of tensor products over arbitrary rings.
Recall that for a ring $R$, an $R$-$R$-bimodule $M$ is called “symmetric” if
$$
  r ⋅ m = m ⋅ r
$$
for all $r ∈ R$ and $m ∈ M$.
When working over a commutative ring $$, people often don’t care in which order they write scalars and elements of modules.
It becomes then impossible to distinguish between left $$-modules, right $$-modules, and symmetric $$-$$-bimodules;
we then only talk about “$$-modules”.
Given any two $$-modules $M$ and $N$, we can therefore use the construction from section 1 (for two left modules), or the construction from this section (for two symmetric bimodules), and both lead to the same $$-module.
When working over $$-algebras
Let $$ be a commutative ring and let $A$ a $$-algebra.
Every left $A$-module is then also a left $$-module by restriction, and every right $A$-module is a right $$-module by restriction.

*

*Suppose that $A$ is commutative, and that $M$ and $N$ are two $A$-modules.
We can then form the tensor products $M ⊗_A N$ and $M ⊗_ N$, both in the sense of section 1.
As a $$-module, $M ⊗_A N$ is a quotient of $M ⊗_ N$, namely by all elements of the form $(a m) ⊗ n - m ⊗ (a n)$ with $m ∈ M$, $n ∈ N$, $a ∈ A$.


*Suppose that $M$ is a right $M$-module and $N$ is a left $A$-module.
We can then form the tensor product $M ⊗_A N$ in the sense of section 3, as well as the tensor product $M ⊗_ N$ in the sense of section 1.
The tensor product $M ⊗_A N$ is a priori only an abelian group, but the $$-module structures on $M$ and $N$ make $M ⊗_A N$ also into a $$-module.
This action of $$ on $M ⊗ N$ is given by
$$
 λ ⋅ (m ⊗ n)
 =
 (λ m) ⊗ n
 =
 m ⊗ (λ m) \,.
$$
The expression $m ⊗ n$ is thus $$-bilinear and $m$ and $n$, but not $A$-bilinear (instead, only $A$-balanced).
As a $$-module, $M ⊗_A N$ is the quotient of $M ⊗_ N$ by all elements of the form $(m a) ⊗ n - m ⊗ (a n)$ with $m ∈ M$, $n ∈ N$, $a ∈ A$.
We see in particular that in any case, the tensor product $M ⊗_A N$ is again a $$-module, and that simple tensors $m ⊗ n$ are $$-bilinear in $m$ and $n$.
So on the underlying $$-level everything is still as used to, it is only on the $A$-level that something new might happen.
4 ♠ Modules over Bialgebras and Hopf algebras
Let $$ be a commutative ring and let $A$ be a $$-algebra.
If $A$ is commutative, then we can tensor together $A$-modules into new $A$-modules.
However, this tensor product over $A$ is not the same as the one over $$ (instead, $M ⊗_A N$ is a quotient of $M ⊗_ N$ as a $$-module).
If $A$ is non-commutative, then we can tensor together a single right $A$-module with a single left $A$-module, but the result will only be a $$-module, not an $A$-module.
Alternatively, we could tensor together $A$-$A$-bimodules into new $A$-$A$-bimodules.
In either case, the resulting tensor product $M ⊗_A N$ is again different from the tensor product $M ⊗_ N$.
So none of our previous constructions solve the following problem:

Question.

*

*Given two left $A$-modules $M$ and $N$, can we extend the tensor product $M ⊗_ N$ into an $A$-module in a functorial way?
(By this functoriality we mean that for homomorphisms of left $A$-modules $f \colon M \to M'$ and $g \colon N \to N'$, the $$-linear map $f⊗ g \colon M ⊗ N \to M' ⊗ N'$ is again a homomorphism of left $A$-modules)

*Can this be done in such a way that the usual natural isomorphisms of $$-modules
$$
 (M ⊗_ N) ⊗_ P ≅ M ⊗_ (N ⊗_ P) \,,
 \quad
 M ⊗_  ≅ M
 \quad
  ⊗_ N ≅ N \,,
$$
are also isomorphisms of $A$-modules?


As a motivation for this question, suppose for a moment that $A$ is the group algebra of some group $G$, i.e., $A = [G]$.
Left modules over $A$ are then the same as representations of the group $G$ over $$.
One of the most basic constructions for new representations is that of tensor products:
if $(V, ρ_V)$ and $(W, ρ_W)$ are two $G$-representation over $$, then one can make the tensor product $V ⊗_ W$ again into a $G$-representation.
In other words, for any two $A$-modules $M$ and $N$, the tensor product $M ⊗_ N$ can again be made into an $A$-module.
The above question tries to understand this process on a more abstract level, for a (possibly) arbitrary $$-algebra $A$.
It turns out that finding an answer to the above question is the same as giving the algebra $A$ the structure of a bialgebra.
Let us be a bit more precise:
Suppose that we find a solution to the first part of the question.
In other words, we find in way to make for any two left $A$-modules $M$ and $N$ the tensor product $M ⊗_ N$ into an $A$-module, in a functorial way.
For every element $a$ of $A$, we can then consider the element
$$
 Δ(a) ≔ a ⋅ (1 ⊗ 1)
$$
of $A ⊗_ A$.
It turns out that the resulting map $Δ$ from $A$ to $A ⊗_ A$ tells us how the action of $A$ on tensor products has to look like:
For any two left $A$-modules $M$ and $N$ and elements $m ∈ M$ and $n ∈ N$, we have unique homomorphisms of $A$-modules
$$
 f \colon A \to M \,
 \quad
 g \colon A \to N
$$
with $f(1) = m$ and $g(1) = n$, given by
$$
 f(a) = am \,,
 \quad
 g(a) = an
$$
for every $a ∈ A$.
The map $f ⊗ g$ from $A ⊗_ A$ to $M ⊗_ N$ is by assumption again a homomorphism of left $A$-modules.
It follows (with Sweedler notation $Δ(a) = \sum_{(a)} a_{(1)} ⊗ a_{(2)}$) that
\begin{align*}
 a ⋅ (m ⊗ n)
 &=
 a ⋅ (f(1) ⊗ g(1))
 \\
 &=
 a ⋅ (f ⊗ g)(1 ⊗ 1)
 \\
 &=
 (f ⊗ g)(a ⋅ (1 ⊗ 1))
 \\
 &=
 (f ⊗ g)\biggl( \sum_{(a)} a_{(1)} ⊗ a_{(2)} \biggr)
 \\
 &=
 \sum_{(a)} f(a_{(1)}) ⊗ g(a_{(2)})
 \\
 &=
 \sum_{(a)} ( a_{(1)} m ) ⊗ ( a_{(2)} n ) \,.
\end{align*}
This calculation tells us that the action of $A$ on any tensor product $M ⊗_ N$ must come from the map $Δ$ in a very specific way.
We can also see that the map $Δ$ must be a homomorphism of $$-algebras from $M ⊗_ N$:

*

*We have $Δ(1) = 1 ⋅ (1 ⊗ 1) = 1 ⊗ 1 = 1_{A ⊗ A}$.


*For all $a, b ∈ A$, we have the chain of equalities
$$
 Δ(a + b)
 =
 (a + b) ⋅ (1 ⊗ 1)
 =
 a ⋅ (1 ⊗ 1) + b ⋅ (1 ⊗ 1)
 =
 Δ(a) + Δ(b) \,.
$$


*For all $λ ∈ $ and $a ∈ A$ we have the chain of equalities
$$
 Δ(λ a)
 =
 (λ a) ⋅ (1 ⊗ 1)
 =
 λ ⋅ (a ⋅ (1 ⊗ 1))
 =
 λ ⋅ Δ(a)
 =
 λ Δ(a) \,.
$$


*The hardest part to check is that $Δ$ is multiplicative.
We have already seen how the module action of $A$ on $A ⊗ A$ has to look like, in terms of $Δ$:
if we denote the multiplication of the algebra $A ⊗_ A$ by $*$, then we have
$$
 a ⋅ t = Δ(a) * t
$$
for all $a ∈ A$ and $t ∈ A ⊗_ A$.
Indeed, in Sweedler notation we have
$$
 Δ(a) = \sum_{(a)} a_{(1)} ⊗ a_{(2)} \,,
 \quad
 t = \sum_{(t)} t_{(1)} ⊗ t_{(2)}
$$
and therefore
\begin{align*}
 a ⋅ t
 &=
 a ⋅ \biggl( \sum_{(t)} t_{(1)} ⊗ t_{(2)} \biggr)
 \\
 &=
 \sum_{(t)} a ⋅ ( t_{(1)} ⊗ t_{(2)} )
 \\
 &=
 \sum_{(t)} \sum_{(a)} ( a_{(1)} t_{(1)} ) ⊗ ( a_{(2)} t_{(2)} )
 \\
 &=
 \sum_{(a)} \sum_{(t)} (a_{(1)} ⊗ a_{(2)}) * (t_{(1)} ⊗ t_{(2)})
 \\
 &=
 \biggl( \sum_{(a)} a_{(1)} ⊗ a_{(2)} \biggr)
 *
 \biggl( \sum_{(t)} t_{(1)} ⊗ t_{(2)} \biggr)
 \\
 &=
 Δ(a) * t \,.
\end{align*}
It follows for any two elements $a$ and $b$ of $A$ that
\begin{align*}
 Δ(ab)
 &=
 (ab) ⋅ (1 ⊗ 1)
 \\
 &=
 a ⋅ (b ⋅ (1 ⊗ 1))
 \\
 &=
 Δ(a) * (Δ(b) * (1 ⊗ 1))
 \\
 &=
 Δ(a) * Δ(b) * 1_{A ⊗ A}
 \\
 &=
 Δ(a) * Δ(b) \,.
\end{align*}
If conversely $Δ$ is any homomorphism of $$-algebras from $A$ to $A ⊗_ A$, then we can use the above formula $a ⋅ (m ⊗ n) = \sum_{(a)} (a_{(1)} m) ⊗ (a_{(2)} n)$ to make $M ⊗_ N$ into an $A$-module, provided both $M$ and $N$ are $A$-modules.
So we see that to solve the first part of the question, we need to choose a homomorphism of $$-algebras from $A$ to $A ⊗_ A$, typically denoted by $Δ$, and then use the very specific formula $a ⋅ (m ⊗ n) = \sum_{(a)} (a_{(1)} m) ⊗ (a_{(2)} n)$.
We don’t have any other choice!
For the second part of the question we can proceed similarly.
I won’t do the complete computations here, but summarize the findings:

*

*The isomorphism of $$-modules $(M ⊗_ N) ⊗_ P ≅ M ⊗_ (N ⊗_ P)$ is an isomorphism of $A$-modules for all left $A$-modules $M$, $N$ and $P$ if and only if $Δ$ is coassociative.


*To make sense of the two isomorphisms $M ⊗_  ≅ M$ and $ ⊗_ N ≅ N$ on the left $A$-modules, we first need to make $$ into a left $A$-module.
Such a left $A$-modules structure on $$ is the same as a homomorphism of $$-algebras $ε$ from $A$ to $\mathrm{End}_() = $.
The two isomorphisms are then equivalent to $ε$ being counital with respect to $Δ$.
In other words, to solve both parts of the question, we need to give $A$ the structure of a $$-bialgebra.
Even more precisely, solutions to the above question are in one-to-one correspondence with bialgebra structures on $A$.

As a summary:
there are different kinds of “tensor product” because not every notion works in every context, and because they serve different purposes:

*

*For modules over commutative rings we have the usual notion of tensor products via bilinearity.


*For modules over arbitrary rings we can still copy the approach via bilinearity, but it doesn’t work very well because this kind of tensor product cannot see the non-commutativity of the ring.
(I don’t think anyone uses this kind of tensor product.)


*We have a rich and useful theory of tensor products over arbitrary rings.
This notion of tensor products generalizes the one for commutative rings, but also behaves differently in its full generality.
(In particular, one needs to tensor a right $R$-module with a left $R$-module, and the result is not again an $R$-module.)


*When working with a $$-algebra $A$, where $$ is a commutative ring, we sometimes want to consider the tensor products over $$, but still as an $A$-module.
To fulfil this wish of ours, we need to endow $A$ with the structure of a bialgebra.
All these kinds of tensor products can occur in praxis, even while studying a specific part of mathematics.
As an example, the following occur in the representation theory of groups over a field $$:

*

*Let $G$ be a group, $H$ a subgroup, and let $M$ be a representation of $H$.
We may regard $M$ as a left $[H]$-module and the group algebra $[G]$ as a $[G]$-$[H]$-bimodule.
We can thus consider the tensor product $[G] ⊗_{[H]} M$ as a left $[G]$-module, and thus as a representation of $G$.
(This is a tensor product as in section 3.)
This is the “natural” way of extending the $H$-representation $M$ to a representation of $G$.


*Let $G$ be a group.
We regard the field $$ as the trivial representation of $G$.
We than have for every other representation $M$ of $G$ the isomorphism $ ⊗_ M ≅ M$.
(This is a tensor product as in section 4, which is also a tensor product in the sense of section 1.)
